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- TL;DR Summary
- Spin-off from previous thread to present math for the kinematic decomposition:

https://www.physicsforums.com/threads/the-paradox-of-relativity-length-contraction.1010138/

In a recent thread, I said that if there was interest, I would post in a separate thread the calculations for the kinematic decomposition of the congruence of worldlines describing the rod in the "rod and hole" relativity paradox discussed in that thread. Since there was interest, I am posting that separate thread.

(Note: Many physicists would probably say that the kinematic decomposition is a graduate level, i.e., "A" level in PF terms, topic. However, for this problem at least, the math is fairly simple and I think it can be presented at an undergraduate, i.e., "I" level. So that is the level I have assigned to this thread.)

The basic definition of the kinematic decomposition is given in this Wikipedia article. The basic steps, at least as I will present them here, are as follows:

(1) Determine the vector field ##U^a## that describes the congruence of worldlines of interest. (The Wikipedia article uses ##X##, but I am using ##U## since that is a more common notation for a 4-velocity field, which is what this is.)

(2) Compute the tensor ##U_{a;b}##, i.e., the covariant derivative of the 1-form ##U_a## corresponding to the vector field ##U^a##. Since we will be working in an inertial frame, the covariant derivative is just the partial derivative, so all we have to compute is ##U_{a,b}##.

(3) Compute the tensor ##K_{ab} = h^m{}_a h^n{}_b U_{m;n}##, i.e., the derivative ##U_{a;b}##, projected orthogonally to ##U##. (The projection tensor ##h## is defined in the Wikipedia article.)

(4) Separate the tensor ##K## (the notation ##K## is my own, to give the entire tensor a name; the literature only names the pieces of it that I'm about to describe) into three pieces: its trace (the expansion scalar ##\theta##), its traceless symmetric part (the shear tensor ##\sigma_{ab}##), and its antisymmetric part (the vorticity tensor ##\omega_{ab}##). It is a mathematical theorem that any 2nd-rank tensor can be fully described by these three pieces, i.e., there is no information left over once we've extracted them. The three pieces are the kinematic decomposition.

Two other notes: first, if the expansion and shear are zero, the congruence can be said to be "Born rigid"; physically, this means the local distances between neighboring worldlines, in the hypersurfaces locally orthogonal to ##U##, are constant. Second, if the vorticity is zero, the congruence can be said to be "hypersurface orthogonal"; physically, this means that we can construct a single set of spacelike hypersurfaces that are everywhere orthogonal to every worldline in the congruence (i.e.,orthogonal not just locally but globally).

With those preliminaries out of the way, here is a brief description of the scenario:

We have a surface along which a rod is moving horizontally in the positive ##x## direction with speed ##v##, until the rod encounters a hole in the surface. We will use the Rindler version of the scenario, in which a trapdoor is placed over the hole and only removed once the entire length of the rod, as seen in the hole/surface rest frame, is over the hole. The instant that occurs, the trapdoor is removed, in such a way that it does not impair the rod free-falling into the hole due to gravity, and in such a way that all parts of the trapdoor are removed at the same time in the hole/surface rest frame, which we will call frame A. We will call that time ##t = 0##.

In the rest frame of the rod before it starts to fall, which we will call frame B, the surface and the hole are moving in the negative ##x'## direction with speed ##v##. Each piece of the rod has a constant ##x'## coordinate in this frame. In this frame, the pieces of the trapdoor are removed at different times, and hence the pieces of the rod start to fall freely at different times. The time ##t' = t_0## at which a piece of the rod at ##x' = x_0## starts falling freely in this frame is given by the condition ##t = 0 = \gamma \left( t' + v x' \right)##, which gives ##t_0 = - v x_0##. This is telling us that the parts of the rod at larger ##x_0## (i.e., further to the right, so they encounter the hole first) start to fall earlier, which is what we expect.

The 4-velocity of each piece of the rod therefore starts out being at rest in this frame (i.e., only a ##t'## component), but then acquires an increasing negative ##z'## component once that piece starts to fall freely. The coordinate acceleration will thus be zero up to time ##t_0## (which, as above, is a function of the position ##x_0## of the piece), and ##- a## after. (We will leave ##a## as an undetermined constant in this analysis, although of course in the scenario as originally posed we expect it to be determined by the Earth's gravity. But we will want to consider various possible values later on.)

The components of ##U^a## are therefore (giving them in order ##t##, ##x##, ##z## and dropping the primes on the coordinates):

$$

U^a = \left( \sqrt{ 1 + a^2 \left( t - t_0 \right)^2 }, 0, - a \left( t - t_0 \right) \right)

$$

Substituting ##t_0 = - v x_0##, and realizing that ##x_0 = x## since each piece of the rod always has the same ##x## coordinate in this frame, we obtain

$$

U^a = \left( \sqrt{ 1 + a^2 \left( t + v x \right)^2 }, 0, - a \left( t + v x \right) \right)

$$

Since these components are in an inertial frame in flat spacetime, so the metric is ##\eta_{ab}##, it is simple to confirm that ##\eta_{ab} U^a U^b = -1##, so ##U## is a timelike unit vector field, as desired.

That completes our first step, and our second step is to compute the tensor ##U_{a,b}##. To do that, we need the 1-form ##U_a##, which since the metric is ##\eta_{ab}## has the same components as ##U^a## except that the first component's sign is changed. So ##U_a## has only ##t## and ##z## components, and the components only depend on ##t## and ##x##; hence, we have four nonzero components of the derivative:

$$

U_{t, t} = \frac{\partial}{\partial t} \left( - \sqrt{ 1 + a^2 \left( t + v x \right)^2 } \right) = - \frac{a^2 \left( t + v x \right)}{\sqrt{ 1 + a^2 \left( t + v x \right)^2 }}

$$

$$

U_{t, x} = \frac{\partial}{\partial x} \left( - \sqrt{ 1 + a^2 \left( t + v x \right)^2 } \right) = - \frac{a^2 v \left( t + v x \right)}{\sqrt{ 1 + a^2 \left( t + v x \right)^2 }}

$$

$$

U_{z, t} = \frac{\partial}{\partial t} \left( - a \left( t + v x \right) \right) = - a

$$

$$

U_{z, x} = \frac{\partial}{\partial x} \left( - a \left( t + v x \right) \right) = - a v

$$

That completes the second step. I'll continue with the two remaining steps in a follow-up post.

(Note: Many physicists would probably say that the kinematic decomposition is a graduate level, i.e., "A" level in PF terms, topic. However, for this problem at least, the math is fairly simple and I think it can be presented at an undergraduate, i.e., "I" level. So that is the level I have assigned to this thread.)

The basic definition of the kinematic decomposition is given in this Wikipedia article. The basic steps, at least as I will present them here, are as follows:

(1) Determine the vector field ##U^a## that describes the congruence of worldlines of interest. (The Wikipedia article uses ##X##, but I am using ##U## since that is a more common notation for a 4-velocity field, which is what this is.)

(2) Compute the tensor ##U_{a;b}##, i.e., the covariant derivative of the 1-form ##U_a## corresponding to the vector field ##U^a##. Since we will be working in an inertial frame, the covariant derivative is just the partial derivative, so all we have to compute is ##U_{a,b}##.

(3) Compute the tensor ##K_{ab} = h^m{}_a h^n{}_b U_{m;n}##, i.e., the derivative ##U_{a;b}##, projected orthogonally to ##U##. (The projection tensor ##h## is defined in the Wikipedia article.)

(4) Separate the tensor ##K## (the notation ##K## is my own, to give the entire tensor a name; the literature only names the pieces of it that I'm about to describe) into three pieces: its trace (the expansion scalar ##\theta##), its traceless symmetric part (the shear tensor ##\sigma_{ab}##), and its antisymmetric part (the vorticity tensor ##\omega_{ab}##). It is a mathematical theorem that any 2nd-rank tensor can be fully described by these three pieces, i.e., there is no information left over once we've extracted them. The three pieces are the kinematic decomposition.

Two other notes: first, if the expansion and shear are zero, the congruence can be said to be "Born rigid"; physically, this means the local distances between neighboring worldlines, in the hypersurfaces locally orthogonal to ##U##, are constant. Second, if the vorticity is zero, the congruence can be said to be "hypersurface orthogonal"; physically, this means that we can construct a single set of spacelike hypersurfaces that are everywhere orthogonal to every worldline in the congruence (i.e.,orthogonal not just locally but globally).

With those preliminaries out of the way, here is a brief description of the scenario:

We have a surface along which a rod is moving horizontally in the positive ##x## direction with speed ##v##, until the rod encounters a hole in the surface. We will use the Rindler version of the scenario, in which a trapdoor is placed over the hole and only removed once the entire length of the rod, as seen in the hole/surface rest frame, is over the hole. The instant that occurs, the trapdoor is removed, in such a way that it does not impair the rod free-falling into the hole due to gravity, and in such a way that all parts of the trapdoor are removed at the same time in the hole/surface rest frame, which we will call frame A. We will call that time ##t = 0##.

In the rest frame of the rod before it starts to fall, which we will call frame B, the surface and the hole are moving in the negative ##x'## direction with speed ##v##. Each piece of the rod has a constant ##x'## coordinate in this frame. In this frame, the pieces of the trapdoor are removed at different times, and hence the pieces of the rod start to fall freely at different times. The time ##t' = t_0## at which a piece of the rod at ##x' = x_0## starts falling freely in this frame is given by the condition ##t = 0 = \gamma \left( t' + v x' \right)##, which gives ##t_0 = - v x_0##. This is telling us that the parts of the rod at larger ##x_0## (i.e., further to the right, so they encounter the hole first) start to fall earlier, which is what we expect.

The 4-velocity of each piece of the rod therefore starts out being at rest in this frame (i.e., only a ##t'## component), but then acquires an increasing negative ##z'## component once that piece starts to fall freely. The coordinate acceleration will thus be zero up to time ##t_0## (which, as above, is a function of the position ##x_0## of the piece), and ##- a## after. (We will leave ##a## as an undetermined constant in this analysis, although of course in the scenario as originally posed we expect it to be determined by the Earth's gravity. But we will want to consider various possible values later on.)

The components of ##U^a## are therefore (giving them in order ##t##, ##x##, ##z## and dropping the primes on the coordinates):

$$

U^a = \left( \sqrt{ 1 + a^2 \left( t - t_0 \right)^2 }, 0, - a \left( t - t_0 \right) \right)

$$

Substituting ##t_0 = - v x_0##, and realizing that ##x_0 = x## since each piece of the rod always has the same ##x## coordinate in this frame, we obtain

$$

U^a = \left( \sqrt{ 1 + a^2 \left( t + v x \right)^2 }, 0, - a \left( t + v x \right) \right)

$$

Since these components are in an inertial frame in flat spacetime, so the metric is ##\eta_{ab}##, it is simple to confirm that ##\eta_{ab} U^a U^b = -1##, so ##U## is a timelike unit vector field, as desired.

That completes our first step, and our second step is to compute the tensor ##U_{a,b}##. To do that, we need the 1-form ##U_a##, which since the metric is ##\eta_{ab}## has the same components as ##U^a## except that the first component's sign is changed. So ##U_a## has only ##t## and ##z## components, and the components only depend on ##t## and ##x##; hence, we have four nonzero components of the derivative:

$$

U_{t, t} = \frac{\partial}{\partial t} \left( - \sqrt{ 1 + a^2 \left( t + v x \right)^2 } \right) = - \frac{a^2 \left( t + v x \right)}{\sqrt{ 1 + a^2 \left( t + v x \right)^2 }}

$$

$$

U_{t, x} = \frac{\partial}{\partial x} \left( - \sqrt{ 1 + a^2 \left( t + v x \right)^2 } \right) = - \frac{a^2 v \left( t + v x \right)}{\sqrt{ 1 + a^2 \left( t + v x \right)^2 }}

$$

$$

U_{z, t} = \frac{\partial}{\partial t} \left( - a \left( t + v x \right) \right) = - a

$$

$$

U_{z, x} = \frac{\partial}{\partial x} \left( - a \left( t + v x \right) \right) = - a v

$$

That completes the second step. I'll continue with the two remaining steps in a follow-up post.

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